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Tju [1.3M]
3 years ago
11

A(n) 60.3 g ball is dropped from a height of 53.7 cm above a spring of negligible mass. The ball compresses the spring to a maxi

mum displacement of 4.68317 cm. The acceleration of gravity is 9.8 m/s 2 . x h Calculate the spring force constant k
Physics
1 answer:
Feliz [49]3 years ago
7 0

Answer:

271.862 N/m

Explanation:

From Hook's Law,

mgh = 1/2ke²............... Equation 1

Where

m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.

Making k the subject of the equation,

k =2mgh/k²....................... Equation 2

Note: The potential energy of the ball is equal to the elastic potential energy of the spring.

Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m

Substitute into equation 2

k = 2(0.0603)(9.8)(0.537)/0.048317²

k = 0.6346696/0.0023345

k = 271.862 N/m

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  • First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
  • After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
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       x = v_{ox} * t = 60.0 m/s * t(s)

  • It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

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  • We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
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  • y(10)= 50 * 9.8 m/s2 = 490.0 m
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